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Radio boxes and ajax / jquery

开发者 https://www.devze.com 2023-03-14 21:53 出处:网络
I have radioboxes <form method=\"post\" id=\"myform\"> <label>A</label> <input type=\"radio\" name=\"formtype\" value = \"1\" checked=\"checked\">

I have radioboxes

<form method="post" id="myform">
        <label>A</label>
        <input type="radio" name="formtype" value = "1" checked="checked">
        <label>B</label>
        <input type="radio" name="formtype" value = "2">
        <label>C</label>
        <input type="radio" name="formtype" value = "3">
        <button type="submit"/>Submit</button>
        <div class = "space"></div>
        </form>

My ajax submission function is:

$.ajax(
        {
        data: 
            {
                type:  /////What do开发者_如何学Python I put HERE <-------

            }, 
        url: 'in.php', 
        complete: function (XMLHttpRequest, textStatus) 
            {
                $('#longurl').val(XMLHttpRequest.responseText);
            }
        });

I am trying to pass in the value of the checked of the 3 radio boxes into something called type, I was wondering what should I put in the arrowed space above (/////What do I put HERE). I was trying to use $('#formtype').val() but it wasn't working.

Thanks


$('#myform input[name="formtype"]:checked').val()

About the :checked selector

What you need to do is, find input elements that have the name attribute of "formtype", then filter to only those that are selected (due to the nature of radioboxes, there will be at most one), then get value of that. The #myform is not strictly necessary, but it speeds things up a bit (because the browser does not need to hunt through the entire page to find the name="formpage" elements).


try this

  $('#myform input[name="formtype"]:checked').val() 
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